Undecidability in Binary Tag Systems and the Post Correspondence Problem for Five Pairs of Words

Transcription

1 Undecidability in Binary Tag Systems and the Post Correspondence Problem for Five Pairs of Words Trlogh Neary Institte of Neroinformatics, University of Zürich and ETH Zürich, Switzerland Abstract Since Cocke and Minsky proved 2-tag systems niversal, they have been extensively sed to prove the niversality of nmeros comptational models. Unfortnately, all known algorithms give niversal 2-tag systems that have a large nmber of symbols. In this work, tag systems with only 2 symbols (the minimm possible) are proved niversal via an intricate constrction showing that they simlate cyclic tag systems. We immediately find applications of or reslt. We redce the halting problem for binary tag systems to the Post correspondence problem for 5 pairs of words. This improves on 7 pairs, the previos bond for ndecidability in this problem. Following or reslt, only the cases for 3 and 4 pairs of words remains open, as the problem is known to be decidable for 2 pairs. In a frther application, we apply the redctions of Vesa Halava and others to show that the matrix mortality problem is ndecidable for sets with six 3 3 matrices and for sets with two matrices. The previos bonds for the ndecidability in this problem was seven 3 3 matrices and two matrices ACM Sbject Classification F.1.2 [Theory of Comptation]: Comptation by Abstract Devices Modes of Comptation Keywords and phrases tag system, Post correspondence problem, ndecidability Digital Object Identifier /LIPIcs.STACS Introdction Introdced by Post [17], tag systems have been sed to prove Tring niversality in nmeros comptational models, inclding some of the simplest known niversal systems [6, 10, 11, 14, 19, 20, 21]. Many niversality reslts rely either on direct simlation of tag systems or on a chain of simlations the leads back to tag systems. Sch relationships between models means that improvements in one model often has applications to many others. The reslts in [23] are a case in point, where an exponential improvement in the time efficiency of tag systems had the domino effect of showing that many of the simplest known models of comptation [6, 10, 11, 14, 19, 20, 21] are in fact polynomial time simlators of Tring machines. Despite being central to the search for simple niversal systems for 50 years, tag systems have not been the sbject of simplification since the sixties. In 1961, Minsky [13] solved Post s longstanding open problem by showing that tag systems, with deletion nmber 6, are niversal. Soon after, Cocke and Minsky [5] proved that tag systems with deletion nmber 2 (2-tag systems) are niversal. Later, Hao Wang [22] showed that 2-tag systems with even shorter instrctions were niversal. The systems of both Wang, and Cocke and Minsky se large alphabets and so have a large nmber of rles. Here we show that tag systems with only 2 symbols, and ths only 2 rles, are niversal. Srprisingly, one of or two rles is trivial. We find immediate applications of or reslt. Using Cook s [6] redction of tag systems to cyclic tag systems, it is a straightforward matter Trlogh Neary; licensed nder Creative Commons License CC-BY 32nd Symposim on Theoretical Aspects of Compter Science (STACS 2015). Editors: Ernst W. Mayr and Nicolas Ollinger; pp Leibniz International Proceedings in Informatics Schloss Dagsthl Leibniz-Zentrm für Informatik, Dagsthl Pblishing, Germany

2 650 Undecidability in Binary Tag Systems and the Post Correspondence Problem to give a binary cyclic tag system program that is niversal and contains only two 1 symbols. We also se or binary tag system constrction to improve the bond for the nmber of pairs of words for which the Post correspondence problem [18] is ndecidable, and the bonds for the simplest sets of matrices for which the mortality problem [16] is ndecidable. The search for the minimm nmber of word pairs for which the Post correspondence problem is ndecidable began in the 1980s [4]. The best reslt ntil now was fond by Matiyasevich and Sénizerges, whose impressive 3-rle semi-the system [12], along with a redction de to Clas [4], showed that the problem is ndecidable for 7 pairs of words. Improving on this ndecidability bond of 7 pairs of words seemed like a challenging problem. In fact, Blondel and Tsitsiklis [2] stated in their srvey The decidability of the intermediate cases (3 n 6) is nknown bt is likely to be difficlt to settle. We give the first improvement on the bond of Matiyasevich and Sénizerges in 17 years: We redce the halting problem for or binary tag system to the Post correspondence problem for 5 pairs of words. This leaves open only the cases for 3 and 4 pairs of words, as the problem is known to be decidable for 2 pairs [7]. A nmber of athors [1, 3, 8, 9, 16], have sed ndecidability bonds for the Post correspondence problem to find simple matrix sets for which the mortality problem is ndecidable. The matrix mortality problem is, given a set of d d integer matrices, decide if the zero matrix can be expressed as a prodct of matrices from the set. Halava et al. [9] proved the mortality problem ndecidable for sets with seven 3 3 matrices, and sing a redction de Cassaigne and Karhmäki [3] they also showed the problem ndecidable for sets with two matrices. Applying the redctions sed in [3, 8] to or new bond, we find that the matrix mortality problem is ndecidable for sets with six 3 3 matrices and for sets with two matrices. In the seqel while simlating cyclic tag systems or binary tag system prodces garbage that grows exponentially dring the simlation, and this reslts in an exponential time simlation overhead. 2 Preliminaries We write c 1 c 2 if a configration c 2 is obtained from c 1 via a single comptation step. We let c 1 t c 2 denote a seqence of t comptation steps. The length of a word w is denoted w, and ɛ denotes the empty word. We let v denote the encoding of v, where v is a symbol or a word. We se the binary modlo operation a = m mod n, where a = m ny, 0 a < n, and a, m, n, and y are integers. 2.1 Tag Systems Definition 1. A tag system consists of a finite alphabet of symbols Σ, a finite set of rles R : Σ Σ and a deletion nmber β N, β 1. The tag systems we consider are deterministic. The comptation of a tag system acts on a word w = w 0 w 1... w w 1 (here w i Σ) which we call the dataword. The entire configration is given by w. In a comptation step, the symbols w 0 w 1... w β 1 are deleted and we apply the rle for w 0, i.e. a rle of the form w 0 w 0,1 w 0,2... w 0,e, by appending the word w 0,1 w 0,2... w 0,e (here w 0,j Σ). A dataword (configration) w is obtained from w via a single comptation step as follows: w 0 w 1... w β... w w 1 w β... w w 1 w 0,1 w 0,2... w 0,e

3 T. Neary 651 where w 0 w 0,1 w 0,2... w 0,e R. A tag system halts if w < β. We se the term rond to describe the w w β or β comptation steps that traverse a word w exactly once. We say a symbol w 0 is read if and only if at the start of a comptation step it is the leftmost symbol (i.e. the rle w 0 w 0,0 w 0,1... w 0,e is applied), and we say a word w = w 0 w 1... w w 1 is entered with shift z < β if w z is the leftmost symbol that is read in w. We let w denote the word obtained by removing the leftmost z symbols of w (i.e. w = [z] [z] w z... w w 1 ) and let w [z] denote the seqence of symbols read dring a single rond on w [z]. So w = w [z] zw z+β w z+2β w z+3β,..., w z+lβ where z + lβ < w. If z < β then w is read when w [z] is entered with shift z and we call w track z of w. A word w has a shift change of 0 s < β [z] if w = yβ s where y N and y > 0. The proof of Lemma 2 is left to the reader. Lemma 2 (shift change). Given a tag system T with deletion nmber β and the word rv Σ, where the word r has a shift change of s and v β, after one rond of T on r entered with shift z the word v is entered with shift (z + s) mod β. 2.2 Cyclic Tag Systems Cyclic tag systems were introdced and proved niversal by Cook [6]. Definition 3. A cyclic tag system C = α 0,..., α p 1 is a list of words α i {0, 1 called appendants. A configration of a cyclic tag system consists of (i) a marker that points to a single appendant α m in C, and (ii) a word w = w 1... w w {0, 1. We call w the dataword. Intitively the list C is a program with the marker pointing to instrction α m. In the initial configration the marker points to appendant α 0 and w is the binary inpt word. Definition 4. A comptation step is deterministic and acts on a configration in one of two ways: If w 1 = 0 then w 1 is deleted and the marker moves to appendant α (m+1 mod p). If w 1 = 1 then w 1 is deleted, the word α m is appended onto the right end of w, and the marker moves to appendant α (m+1 mod p). A cyclic tag system completes its comptation if (i) the dataword is the empty word or (ii) it enters a repeating seqence of configrations. As an example we give first 5 steps of the cyclic tag system C = 001, 01, 11 on the inpt word 101. In each configration C is given on the left with the marked appendant highlighted in bold font. 001, 01, , 01, , 01, , 01, , 01, , 01, The Halting Problem for Binary Tag Systems Definition 5 (Halting problem for binary tag systems). Given a 2-symbol tag system T with deletion nmber β and a dataword w, does T prodce a seqence of comptation steps of the form w w where w < β? Theorem 6. The halting problem for binary tag systems is ndecidable. The proof of Theorem 6 proceeds in two stages. First we constrct a non-halting binary tag system T C that simlates an arbitrary cyclic tag system C (Sections 3.1 and 3.2). Following this in Lemma 9 we show how to modify or constrction so that when simlating the cyclic tag system in [15] or system halts if an only if it is simlating a halting Tring machine. S TAC S

4 652 Undecidability in Binary Tag Systems and the Post Correspondence Problem Table 1 Table defining. In the middle colmn is the seqence of symbols (track) read in when the object in the left colmn is entered with shift (β 4m) mod β, where β = 4p is the deletion nmber, α m is a cyclic tag system appendant, and σ i is a binary word where σ i = bbcbb if σ i = 0, and σ i = bbbcb if σ i = 1. Also σ 1 is the word σ 1 with its leftmost symbol removed. Object Track read in Vales for m and α m ɛ = bbbb [(β 4m) mod β] = cs 0 m < p [β 1] = cbbb(bcbbb)p 1 c s 5p+1 m = 0 [β 4m 1] = cs 0 < m < p 0 = bbbb [β 2] = cbbb(bcbbb)p 1 c s 5p+1 m = 0 [β 4m 2] = cs 0 < m < p 1 = bbbb [β 3] = cbbb(bcbbb)p 1 c s 5p+1 α 0 = ɛ, m = 0 [β 3] σ1 σ 2... σ v c s 5v+1 α 0 = σ 1σ 2... σ v for v > 0, m = 0 [β 4m 3] = σ1 σ 2... σ v c s 5v α m = σ 1σ 2... σ v for 0 < m < p 3.1 Binary Tag System T C and Its Encoding Here we give a binary tag system T C that simlates the comptation of an arbitrary C = α 0,..., α p 1. The deletion nmber of T C is β = 4p, its alphabet is {b, c, and its rles are of the form b b and c, where {b, c. The binary word encodes the entire program of C and is defined by Table 1, where = βs, s 5(max(p, r)) and r is the length of the longest appendant in C. See Section for an example of how Table 1 is sed to give. The cyclic tag system symbols 0 and 1 are encoded as the binary words 0 = bbbb and 1 = bbbb respectively. We refer to 0 and 1 as objects. Definition 7 (Inpt to T C ). An arbitrary inpt dataword w 1 w 2... w n {0, 1 to a cyclic tag system is encoded as the T C inpt dataword w 1 w 2... w n. Dring the simlation we make se of an extra garbage object: the binary word ɛ = bbbb. The cyclic tag system configration (α 0, α 1... α m 1 α m α m+1... α p 1 w 1 w 2... w l ) is encoded as w 1 { ɛ p, w 2 { ɛ p, w 3... { ɛ p, w l { ɛ p, (1) [(β 4m) mod β] where w 1 [(β 4m) mod β] denotes the word given by an object w 1 { 0, 1 with its leftmost [β 4m) mod β] symbols deleted. This implies that w 1 is entered with the shift [(β 4m) mod β] and this shift vale records that the crrently marked cyclic tag system appendant is α m. If a sbword in the dataword of T C does not form part of one of the three objects 0, 1 and ɛ we will refer to this sbword as a garbage object. So words of the form { ɛ, in Eqation (1) consist only of garbage objects. 3.2 The Simlation Algorithm The seqence of symbols that is read in an object is determined by the shift vale with which it is entered (see Section 2.1). So in the simlation the shift vale is sed for algorithm

5 T. Neary 653 (i) 1 a 2 a 3... a l s a 2 a 3... a l σ 1 σ 2... σ v s 5v [β 4m] [(β 4(m+1)) mod β] (ii) 0 a 2 a 3... a l s a 2 a 3... a l s [β 4m] [(β 4(m+1)) mod β] (iii) (iv) ɛ a 2 a 3... a l s a 2 a 3... a l s [β 4m] [(β 4(m+1)) mod β] a 2 a 3... a l s a 2 a 3... a l s [(β 4m) mod β] [(β 4m) mod β] Figre 1 Objects 1, 0, ɛ and being read when entered with shift [β 4m]. Above s denotes the s comptation steps that read each object, a i {, 0, 1, ɛ, β = 4p is the deletion nmber, and p is the length of C s program. In (i), (ii) and (iii) 0 < m < p and in (iv) 0 m < p. On the left is the dataword before the object is read and on the right is the dataword after the object has been read. After 1, 0 or ɛ is read the adjacent object a 2 is entered with shift [β 4(m + 1)], and in (iv) after is read a 2 is entered with shift [(β 4m) mod β]. Reading the objects 0, ɛ or appends s, and reading the object 1 appends the encoding of cyclic tag system appendant α m = σ 1σ 2... σ v. control flow. Figres 1 and 2 give a high level view of reading each of the for objects that appear in Eqation (1) and this view incldes the shift change that occrs from reading each object. Objects 1, 0 and ɛ have length 1 = 0 = ɛ = βs + 4 and so have a shift change of β 4 (see end of Section 2.1). So when these objects are entered with shift [(β 4m) mod β] the adjacent object a 2 is entered with shift [(β 4(m + 1)) mod β] (as shown in Figres 1 and 2). When reading 1 or 0 this shift change of β 4 simlates the appendant marker moving from appendant α m to α ((m+1) mod p). Recall that β = 4p and so reading p of the 0 and 1 objects has a shift change of 0 (since 0 = p(β 4) mod β). This shift change of 0 means that the encoding of the marked appendant retrns to its original vale after reading p objects, correctly simlating that the appendant marker beginning at α m has moved throgh the entire length p circlar program of C and retrned to it original position of marking α m. The garbage objects ɛ and in Eqation (1) appear only in garbage words of the form { ɛ p, which have a shift change of 0 (since ɛ p = (1 + ps)β and = βs) and so have no effect on the shift vale when entering sbseqent objects (see Lemma 2). In Figres 1 and 2 we see that reading an 1 simlates reading a 1 by appending the encoding of the marked appendant α m = σ 1 σ 2... σ v and reading an 0 simlates reading a 0 by appending a garbage word from { ɛ p, to simlate that nothing is appended. In Figres 1 and 2 when the garbage objects ɛ and are read they append garbage words from { ɛ p, that have no effect on the simlation. To help the reader, in Section 3.3 we define for a specific example and se this vale to give an example of or algorithm simlating a cyclic tag system comptation step. Here we explain how Table 1 defines so that when each object is read it appends the appendants shown in Figres 1 and 2. Recall that a track w is the seqence of symbols read in a word w [z] when it is entered with shift z (see Section 2.1). Table 1 defines by giving (in the middle colmn) each possible track in (for detailed example see Section 3.3.1). Each track is read when the object in the left colmn of Table 1 is entered with shift [(β 4m) mod β]. To see this note that the nmber of b symbols that proceed the word in each object cases a shift change which in trn cases the track given in middle colmn of Table 1 to be read. For example the word bbb at the left end of 1 = bbbb has a shift change vale of β 3 (see Section 2.1) and ths when 1 is entered with shift [(β 4m) mod β] we enter with shift [β 4m 3]. So when m > 0 and we read an 1, we see from row 8 of Table 1 that track = σ [β 4m 3] 1 σ 2... σ v c s 5v is read. Applying the rles S TAC S

6 654 Undecidability in Binary Tag Systems and the Post Correspondence Problem (i) 1 a 2 a 3... a l s+1 a 2 a 3... a l σ 1 σ 2... σ v s 5v+1 [0] [β 4] (ii) 0 a 2 a 3... a l s+1 a 2 a 3... a l ɛ p s 5p+1 [0] [β 4] (iii) ɛ [0] a 2 a 3... a l s+1 a 2 a 3... a l ɛ p s 5p+1 [β 4] Figre 2 Objects 1, 0 and ɛ being read when entered with shift 0. Above s+1 denotes the s + 1 comptation steps that read each object when entered with shift 0. Here a i {, 0, 1, ɛ, β = 4p is the deletion nmber, and p is the length of C s program. On the left is the dataword before the object is read and on the right is the dataword after the object has been read. After 1, 0 or ɛ is read the adjacent object a 2 is entered with shift [β 4]. Reading the objects 0 or ɛ appends ɛ p s 5p+1, and reading the object 1 appends the encoding of cyclic tag system appendant α 0 = σ 1σ 2... σ v. b b and c when reading this track appends the appendant σ 1 σ 2... σ v c s 5v as shown in Figre 1 (i). For m > 0 the entire seqence of symbol read in 1, 0, and ɛ are respectively given by the tracks in rows 3, 5 and 8 of Table 1. One can see that applying the rles b b and c to these tracks appends the correct appendant for each object read in Figre 1. For m = 0 we have a special case where to get the entire seqence of symbols read in each object we prepend an extra b to the tracks given in rows 2, 4, 6 and 7 of Table 1 and now applying the rles b b and c to this seqence appends the appendants shown in Figre 2. For example, to give the seqence read in 1 when m = 0, we prepend b to track = σ [β 3] 1 σ 2... σ v c s 5v+1 (row 7 of Table 1) to give the seqence b σ 1 σ 2... σ v c s 5v+1 = σ 1 σ 2... σ v c s 5v+1 which appends the word σ 1 σ 2... σ v c s 5v+1 as shown in Figre 2 (i). Finally, in row 1 of Table 1 are the tracks that give the seqence of symbols read in garbage objects for all vales of m Tag system T C Simlating an Arbitrary Comptation Step of C Eqation (2) gives an arbitrary comptation step of cyclic tag system C, where w = σ 1... σ v if w 1 = 1 and α m = σ 1 σ 2... σ v, and w = ɛ if w 1 = 0. α 0,...α α m α m+1... α p 1 w 1 w 2... w l α 0,... α m α m+1... α p 1 w 2... w l w (2) Lemma 8 essentially states that T C simlates the arbitrary comptation step given in Eqation (2). In Lemma 8, Eqations (3) and (4) respectively encode the left and right configrations in Eqation (2) (see Eqation (1)). Lemma 8 (T C simlates an arbitrary comptation step of C). Given a dataword of the form w 1 { ɛ p, w 2 { ɛ p, w 3... { ɛ p, w l { ɛ p, (3) [(β 4m) mod β] where w i {0, 1, T C reads the word w 1 { ɛ p, to give a dataword of the form [(β 4m) mod β] w 2 { ɛ p, w 3... { ɛ p, w l { ɛ p, w { ɛ p, (4) [(β 4(m+1)) mod β] where w = σ 1 σ 2... σ v if w 1 = 1 and α m = σ 1 σ 2... σ v, and w = ɛ if w 1 = 0.

7 T. Neary 655 Proof. We se Figres 1 and 2 to verify that given Configration (3), T C prodces Configration (4). Following this we discss the correctness of Figres 1 and 2. From (i) and (ii) of Figres 1 and 2 there are 4 possible cases for reading w 1, where each case is determined by the vale of w 1 { 1, 0 and the shift ([0] or [β 4m] for m > 0) with which it is entered. The techniqe for verifying that T C prodces Configration (4) from Configration (3) is similar for all 4 cases and so we will only go throgh one case. We choose the case where w 1 = 1 and is entered with shift [β 4m] for m > 0. From Figre 1 (i) when we read w 1 = 1 with shift [β 4m] in Configration (3) we get { ɛ p, w 2 { ɛ p, w 3... { ɛ p, w l { ɛ p, σ 1 σ 2... σ v s 5v (5) [(β 4(m+1) mod β] In Configration (5) the word { ɛ p, is entered with shift [(β 4(m + 1) mod β]. Each ɛ has a shift change of β 4 and each has a shift change of 0, and so as we read the word { ɛ p, the ɛ and objects are entered with shifts of the form [(β 4m ) mod β] where 0 m < p. So the cases for reading the objects in the word { ɛ p, are given by (iii) in Figres 1 and 2 and (iv) in Figre 1. Ths when we read the word { ɛ p, at the left end of Configration (5) we append a word of the form { ɛ p, as shown in Configration (6). Recall that words of the form { ɛ p, have a shift change of [0] and so when we enter { ɛ p, with shift [β 4(m + 1) mod β] as shown in Configration (5) we also enter the adjacent object w 2 with shift [β 4(m + 1) mod β] as shown in Configration (6). w 2 { ɛ p, w 3... { ɛ p, w l { ɛ p, σ 1... σ v s 5v { ɛ p, (6) [(β 4(m+1) mod β] Configration (6) is of the form of Configration (4) for the case where w 1 = 1 and is entered with shift [(β 4m) mod β] for m > 0. So for this case we have shown sing Figres 1 and 2 that given Configration (3) T C prodces Configration (4). The correctness of Figres 1 and 2 can be verified by showing that on each line of these figres, reading the leftmost object in the configration on the left prodces the configration on the right. This is achieved by showing that when we enter the adjacent object a 2 with the correct shift and also that the correct word gets appended at the end of the dataword. The shift vale with which a 2 is entered follows immediately from Lemma 2 and the length of the object being read. We direct the reader to paragraph 2 of Section 3.2 where we explain how Table 1 defines so that when each object is read it appends the appendants shown in Figres 1 and 2. Or first main Theorem (Theorem 6) follows from Lemma 9. For tag system T C in Lemma 9 we restrict the type of inpt allowed as this lets s simlate T C in Theorem 11 when proving the Post correspondence problem ndecidable for 5 pairs of words. Lemma 9. Let T C be an arbitrary binary tag system with deletion nmber β, alphabet {b, c and rles of the form b b and c l b ( i {b, c). The halting problem is ndecidable for tag systems of the form of T C when given β 1 β... l b as inpt. Proof. In [15] the cyclic tag system C simlates the comptation of an arbitrary Tring machine and appends an appendant (which we will call α h ) encoding the Tring machine halt state if and only if the simlated Tring machine halts. So given the cyclic tag system C, its inpt w 1 w 2... w n, and an appendant α h, we constrct a binary tag system T C that takes β 1 β... l b as inpt and simlates C on inpt w halting if an only if C appends α h. It then follows that halting problem is ndecidable for tag systems of the form of T C when given β 1 β... l b as inpt. S TAC S

8 656 Undecidability in Binary Tag Systems and the Post Correspondence Problem Table 2 Table defining for tag system in Lemma 9. In the middle colmn is the seqence of symbols (or track) read in when the object in the left colmn is entered with shift [(β 10m + 1) mod β], where β = 10p is the deletion nmber, α m is a cyclic tag system appendant, ɛ = b 4 cb 6, and σ i = b 6 cb 4 if σ i = 0, and σ i = b 8 cb 2 if σ i = 1. Also ɛ and σ 1 are the words ɛ and σ 1 with their leftmost symbol removed. Object Track read in Vales for m and α m inpt track [β 1] bβ 2 w 1... w n s 11(n+p) β+2 ( ɛ ) p [(β 10m+1) mod β] cs 0 m < p ɛ = b 4 b 6 ( ɛ ) p 1 c s 11p+1 [β 3] m = 0 [β 10m 3] ( ɛ ) p c s 11p 0 < m < p 0 = b 6 b 4 ( ɛ ) p 1 c s 11p+1 [β 5] m = 0 [β 10m 5] ( ɛ ) p c s 11p 0 < m < p 1 = b 8 b 2 [β 7] σ1 σ 2... σ v c s 11v+1 α 0 = σ 1... σ v, m = 0 halting tracks = [β 10m 7] σ1 σ 2... σ v c s 11v = [β 10h 7] bs 1 = [2i] bs α m = σ 1... σ v, 0 < m < p, m h i = {0, 1, 2, 3,... β 2 2 We obtain T C by modifying tag system T C from Section 3.1. In T C, is defined by Table 2 and we have 0 = b 6 b 4, ɛ = b 4 b 6, 1 = b 8 b 2. The deletion nmber is now β = 10p, and = βs with s = 11(max(p+n+β 2, r)). On inpt β 1 β... l b tag system T C reads track and from row 1 of Table 2 this appends [β 1] bβ 2 w 1 w 2... w n s 11(n+p) β+2 ( ɛ ) p. So T C begins its comptation by appending the encoding of the inpt to C. Reading β 1 β... l b, which has a shift change of β 1, cases s to enter the encoded dataword with a shift of β 1 and this means that we can enter garbage objects with shift β 1 and read the track that appends the encoded inpt. To avoid this we append the word b β 2 to the left of w 1 and since b β 2 has a shift change of 2 the encoded inpt is entered with shift (instead of [β 1]). After reading this b β 2 T C begins the simlation of C on inpt w, making se of the same algorithm as T C. Since we enter the encoding of the inpt word w with shift and objects 0 = b 6 b 4 and 1 = b 8 b 2 now have a shift change of β 10, T C enters objects with shifts of the form [(β 10m + 1)) mod β]. This means that when reading garbage object and objects ɛ, 0, and 1, we enter with shifts of [(β 10m + 1)) mod β], [(β 10m 3)) mod β], [(β 10m 5)) mod β] and [(β 10m 7)) mod β] respectively. This gives the shift vales for the tracks in Table 2. By comparing the tracks in Tables 1 and 2 we can see that the when objects in T C are read they append similar seqences of objects to those in T C. So the comptation of T C proceeds in the same manner as the comptation of T C. However, if C appends α h then we enter 1 with shift [β 6h] and track = [β 6h 4] bcs 1 is read appending the word b s 1. When b s 1 is read dring the next traversal of the dataword the single b in this word cases a shift change of β 1 which means that all sbwords in the dataword of T C will now be entered with an even valed shift. From row 10 of Table 1 all even valed tracks in append only b symbols and so after one frther traversal the dataword consists entirely of b symbols. After this the rle b b, which appends one b and deletes β symbols, is repeated ntil the nmber of symbols is < β and the comptation halts.

9 T. Neary 657 Table 3 Table defining for the cyclic tag system C = 10, 0. In the middle colmn is the seqence of symbols (track) read in when the object in the left colmn is entered with shift (8 4m) mod 8, where α m is a cyclic tag system appendant, 1 = bbbcb, 0 = bbcbb and 0 = bcbb. Object Track read in Vales for m and α m ɛ = bbbb [(8 4m) mod 8] = c10 0 m < 2 [7] = cbbb(bcbbb)c m = 0 [3] = c10 m = 1 0 = bbbb [6] = cbbb(bcbbb)c m = 0 [2] = c10 m = 1 1 = bbbb = 0 c 6 α [5] 0 = 0 = 0 1 α 1 = Example Simlation for T C Using Table 1 to define To help explain how Table 1 is sed to define, we take the example of defining for the cyclic tag system program C = 0, 01. The vale for is given in Eqation (7), where to improve readability, we have split into two eqal length sbwords and with a space between every forth symbol. = = cbcc cbcc cbcc ccbb cccc cbbb cbcc cbbb cbcc ccbb = cbcc cccc cbcc ccbb cbcc ccbb cccc ccbb cbcc cccc (7) From Section 3.1 when C = 0, 01 we have p = 2, s 10, β = 8, = 80, α 0 = 0 and α 1 = 01. By sbstitting these vales into Table 1 we get Table 3 which defines for or tag system that simlates C = 0, 01. Table 3 defines the word as a series of tracks. Recall from Section 2.1 that a track w [z] = w zw z+β w z+2β,..., w z+lβ in a word w is the seqence of symbol read in that word when it is entered with shift z. Here β = 8 and so for m = 0, row 1 of Table 3 defines track [0] = = c 10, which is shown in bold below. = cbcc cbcc cbcc ccbb cccc cbbb cbcc cbbb cbcc ccbb = cbcc cccc cbcc ccbb cbcc ccbb cccc ccbb cbcc cccc Taking row 2 of Table 3 gives [7] = = cbbb(bcbbb)c which again is given in bold immediately below. = cbcc cbcc cbcc ccbb cccc cbbb cbcc cbbb cbcc ccbb = cbcc cccc cbcc ccbb cbcc ccbb cccc ccbb cbcc cccc Rows 3 to 7 of Table 3 give the tracks that complete the definition of for or tag system that simlates the cyclic tag system C = 10, 0. S TAC S

10 658 Undecidability in Binary Tag Systems and the Post Correspondence Problem Simlating a Comptation Step with T C. In this section we give the low level details of or simlation algorithm for T C by simlating the first comptation step (0, , 01 10) of cyclic tag system C = 0, 01 on the inpt dataword 11. The inpt dataword 11 to C is encoded via Definition 7 as the tag system dataword 1 1 and sing 1 = bbbb and Eqation (7) this can be rewritten as { 1 { bbb cbcc cbcc cbcc ccbb cccc {{ cbbb cbcc cbbb cbcc ccbb b 1 (8) Becase the dataword is qite long we only give the left end of the dataword as b and c symbols and se higher level objects on the right. In configration (8) the leftmost symbol is b and so we apply the rle b b by appending b and deleting the leftmost 8 symbols from the dataword to give bcc cbcc ccbb cccc cbbb {{ cbcc cbbb cbcc ccbb b 1 b (9) [5] Above is entered with shift 5 and so we begin reading track from Table 3. We apply [5] the rles b b and c of T C to give the next for comptation steps cbb cccc cbbb cbcc cbbb cbcc ccbb b 1 bb bbb cbcc cbbb cbcc ccbb b 1 bb bbb cbcc ccbb b 1 bbb cbb b 1 bbbb {{ 0 Dring the first 5 comptation steps the word 0 = bbbb was appended to the dataword. Below we give a rewritten form of the configration immediately above where bbbb is replaced with 0 and is replaced with its vale from Eqation (7). We also give the next 5 comptation steps cbb cbcc cccc cbcc ccbb cbcc {{ ccbb cccc ccbb cbcc cccc b ccc b Below we have rewritten the configration given immediately above and given the last configration in this simlated comptation step. ccc b { 1 { bbb cbcc cbcc cbcc ccbb cccc {{ cbbb cbcc cbbb cbcc ccbb b 0 5 (10) 1 [4] {{ bcc cbcc cbcc ccbb cccc {{ cbbb cbcc cbbb cbcc ccbb b 0 6 (11)

11 T. Neary 659 In Eqation (11) above we have finished reading the leftmost 1 in the dataword and completed or simlation of the comptation step (0, , 01 10). The word 0 6 was appended simlating appendant α 0 = 0 from the program C = 0, 01 was appended. From Section 3.1 the sbwords in 6 are considered garbage objects and these sbwords have no effect on the comptation (see first paragraph of Section 3.2). Also, in Eqation (11) we see that the next 1 is entered with shift 4 which encodes that second appendant α 1 = 01 in the program C = 0, 01 is now marked. To see this recall that entering an object with a shift of (β 4m) mod β encodes that appendant α m is marked and since β = 8 we get a shift of 4 when m = 1 meaning α 1 is marked. The dataword in Eqation (11) is of the form given in Eqation (1) and is ready to begin simlation of the next comptations step. 4 The Post Correspondence Problem for 5 Pairs of Words In Theorem 11 we redce the halting problem for the binary tag system given in Lemma 9 to the Post correspondence problem for 5 pairs of words. Definition 10 (Post correspondence problem). Given a set of pairs of words {(r i, v i ) r i, v i Σ, i {0, 1, 2..., n where Σ is a finite alphabet, determine whether or not there exists a non-empty seqence i 1, i 2,... i l, {0, 1, 2..., n sch that r i1 r i2... r il = v i1 v i2... v il. Theorem 11. The Post correspondence problem is ndecidable for 5 pairs of words. Proof. We redce the halting problem for the binary tag systems T C Post correspondence problem instance given by the 5 pairs of words P = {(1, l 10), (10 β 1, 110), (10 β 1, 0), (1, 0), (10 β 1111, 1111) in Lemma 9 to the where ɛ is the empty word and i {b, c. The symbols b and c in T C are encoded as b = 10 β 1 and c = 1 respectively, where β is the deletion nmber of T C. Let r = r i 1 r i2... r il and v = v i1 v i2... v il, where each (r i, v i ) P and r is a prefix of v. We will call the pair (r, v) a configration of P. An arbitrary dataword x 0 x 1... x q b {b, c b is encoded by a P configration of the form (r, v) = (r, r x 0 x 1... x q 10 β ) (12) In each configration (r, v), the nmatched part of v (i.e. x 0 x 1... x q 10 β ) encodes the crrent dataword of T C. Starting from the pair (1, l 10), if 0 = c we add the pair (1, 0) and this matches c = 1 simlating the deletion of 0. If, on the other hand, 0 = b we add the pair (10 β 1, 0) and this matches b = 10 β 1 simlating the deletion of 0. So after matching 0 we have (1 0, l 100). We match β 1 encoded T C symbols in this way to give (r, v) = ( β 2, l 10 β ). The configration is now of the form given in Eqation (12) and the nmatched seqence β 1... l 10 β in v encodes the inpt dataword to T C in Lemma 9. A step of T C on an arbitrary dataword x 0x 1... x q b is of one the two forms: cx 1... x q b x β 1... x q b 1... l b (13) bx 1... x q b x β 1... x q bb (14) These comptation steps are simlated as follows: In Eqation (12), if x 0 = c then x 0 = 1 and we add the pair (1, l 10) to simlate the T C rle c 0... l b, and S TAC S

12 660 Undecidability in Binary Tag Systems and the Post Correspondence Problem this gives (r1, r1 x 1... x q 10 β l 10). In Eqation (12), if x 0 = b then x 0 = 10 β 1 and we add the pair (10 β 1, 110) to simlate the T C rle b b, giving (r10 β 1, r10 β 1 x 1... x q 10 β 110). In both cases we simlate the deletion of a frther β 1 tag system symbols as we did in the previos paragraph to complete the simlated comptation step. So if x 0 = c this gives (r1 x 1... x β 1, r1 x 1... x q 10 β l 10 β ), with the nmatched part in this pair encoding the new dataword on the right of Eqation (13). Or, if x 0 = b we get (r10 β 1 x 1... x β 1, r10 β 1 x 1... x q 10 β 110 β ), with the nmatched part in this pair encoding the new dataword on the right of Eqation (14). The simlated comptation step for both cases is now complete. Now we show that P simlates T C halting with a matching pair of words. Recall that = sβ where we can choose s to be any natral nmber greater than some constant (see Section 3.1). We choose s to be of the form s = x(β 1) + 1 and so the inpt dataword β 1... l b to T C has length xβ(β 1)+1 and the rles either increase its length by xβ(β 1) (rle c 0... l b) or decrease it by β 1 (rle b b), which means all datawords of T C have length y(β 1) + 1, where y N. From Lemma 9, T C halts when the length of its dataword (which has the form b ) is less than the deletion nmber β. So, when T C halts we have y(β 1) + 1 < β which means the dataword is a single b. From Eqation (12), this is encoded as the configration (r, v) = (r, r10 β ). By appending the pair (10 β 1111, 1111) to (r, v), we get the matching pair (r10 β 1111, r10 β 1111) when T C halts. To complete or proof we show that choices that do not follow the simlation as described above leads to a mismatch. We mst have (1, l 10) as the leftmost pair as ptting any other pair from P on the left will not give a match. Now recall that the pair (1, l 10) encodes the initial configration (or inpt) for T C. So now we show that given the encoding of an arbitrary configration any choice that does not follow the simlation leads to a mismatch. From Eqation (12) an arbitrary configration has the form (r, r x 0 x 1... x q 10 β ). If x 0 = b then x 0 = 10 β 1 and we cannot choose either of the pairs (1, l 10) or (1, 0) as they will allow no frther matches, the pair (10 β 1111, 1111) cannot be chosen as for consective 1s do not appear in the encoding (this is becase in we cannot have 2 encoded c symbols ( c = 1) next to each other). The pair (10 β 1, 0) appends a 0 onto the right seqence to give 10 β+1 which cannot be matched as it is only possible to match 0 seqences of the from 10 β 1. Similar argments are sed for the case of x 0 = c and so we do not repeat them. After we have matched x 0 or simlation algorithm reqires that we simlate the deletion of a frther β 1 symbols. If we deviate from the simlation either we find almost immediately that no more matches are possible or we end p with a seqence of zeros that does not have the form 10 β 1 and so cannot be matched. After simlating the deletion of β 1 symbols we have completed the simlation of an arbitrary comptation step and arrived at an encoded configration of the form given by Eqation (12). So it is not possible to find a match in P if we do not follow the simlation described above. Therefore, P has a matching seqence if and only if T C halts. By applying the redctions in [8] and [3] to P in Theorem 11 we get Corollary 12. Corollary 12. The matrix mortality problem is ndecidable for sets with six 3 3 matrices and for sets with two matrices. Acknowledgements: This work is spported by Swiss National Science Fondation grant nmber I wold like to thank Matthew Cook, Damien Woods, Vesa Halava, and Mika Hirvensalo for their comments and discssions.

13 T. Neary 661 References 1 Vincent D. Blondel and John N. Tsitsiklis. When is a pair of matrices mortal? Information Processing Letters, 63(5): , Vincent D. Blondel and John N. Tsitsiklis. A srvey of comptational complexity reslts in systems and control. Atomatica, 36(9): , Jlien Cassaigne and Jhani Karhmäki. Examples of ndecidable problems for 2-generator matrix semigrop. TCS, 204(1-2):29 34, Volker Clas. Some remarks on PCP(k) and related problems. Bll. EATCS, 12:54 61, John Cocke and Marvin Minsky. Universality of tag systems with P = 2. Jornal of the ACM, 11(1):15 20, Matthew Cook. Universality in elementary celllar atomata. Complex Systems, 15(1):1 40, Andrzej Ehrenfecht, Jhani Karhmäki, and Grzegorz Rozenberg. The (generalized) Post correspondence problem with lists consisting of two words is decidable. TCS, 21(2): , Vesa Halava and Tero Harj. Mortality in matrix semigrops. American Mathematical Monthly, 108(7): , Vesa Halava, Tero Harj, and Mika Hirvensalo. Undecidability bonds for integer matrices sing Clas instances. IJFCS, 18(5): , Tero Harj and Marice Margenstern. Splicing systems for niversal Tring machines. In DNA 10, volme 3384 of LNCS, pages Springer, Kristian Lindgren and Mats G. Nordahl. Universal comptation in simple one-dimensional celllar atomata. Complex Systems, 4(3): , Yri Matiyasevich and Gérad Sénizerges. Decision problems for semi-the systems with a few rles. TCS, 330(1): (An earlier version appeared in 11th Annal IEEE Symposim on Logic in Compter Science, LICS 1996".), Marvin Minsky. Recrsive nsolvability of Post s problem of tag" and other topics in theory of Tring machines. Annals of Mathematics, 74(3): , Marvin Minsky. Size and strctre of niversal Tring machines sing tag systems. In Recrsive Fnction Theory: Proceedings, Symposim in Pre Mathematics, volme 5, pages , Provelence, AMS. 15 Trlogh Neary and Damien Woods. P-completeness of celllar atomaton Rle 110. In ICALP 2006, Part I, volme 4051 of LNCS, pages Springer, Michael S. Paterson. Unsolvability in 3 3 matrices. Stdies in Applied Mathematics, 49(1): , Emil L. Post. Formal redctions of the general combinatorial decision problem. American Jornal of Mathematics, 65(2): , Emil L. Post. A variant of a recrsively nsolvable problem. Blletin of The American Mathematical Society, 52: , Raphael M. Robinson. Undecidability and nonperiodicity for tilings of the plane. Inventiones Mathematicae, 12(3): , Yrii Rogozhin. Small niversal Tring machines. TCS, 168(2): , Pal Rothemnd. A DNA and restriction enzyme implementation of Tring machines. In DNA Based Compters, volme 27 of DIMACS, pages AMS, Hao Wang. Tag systems and lag systems. Mathematical Annals, 152:65 74, Damien Woods and Trlogh Neary. On the time complexity of 2-tag systems and small niversal Tring machines. In 47 th Annal IEEE Symposim on Fondations of Compter Science, pages , S TAC S